list - 用 '.' 将一个字符串切割成一个字符串列表

Question

我必须编写一个 cutString::String -> [String] 函数，该函数必须将一个字符串剪切成一个字符串列表。它应该在'。字符。

例如:

cutString "this.is.a.string" == ["this","is","a","string"]

cutString "" == [""]

cutString ".." == ["",""]

cutString ".this.is.a.string." == ["","this","is","a","string",""]

到目前为止，我有这个:

cutString [] = []
cutString [x]
  | x == '.' = []
  | otherwise = [[x]]
cutString (x:xs) = cutString [x] ++ cutString xs

这只省略了句点(这部分是可取的)，但也逐个字符地切割整个字符串。

像这样:

["t","e","s","t","t","e","s","t","2"]

Answer 1

您只是在查看第一个字符是否是一个点，表示具有一个元素的字符串。您应该检查每个元素。如果第一个字符是点，我们会生成一个空字符串，然后是通过递归生成的其余组。如果字符不是点，我们将递归调用的第一个子列表放在前面，例如:

cutString :: String -> [String]
cutString [] = [""]
cutString ('.':xs) = "" : cutString xs
cutString (x:xs) = let <strong>~(y:ys)</strong> = cutString xs in <strong>(x:y)</strong> : ys